Chemistry MCQs for NEET — Practice Questions with Answers

(c) As the no. of particles is highest for the 1(M) BaCl₂ after complete ionisation, therefore, elevation of boiling point will be highest for this solution.

(c)

 $$\begin{gathered} 0.18={\mathrm{K}}_{\mathrm{b}}\times \frac{2\times 1000}{60\times 98}, {\mathrm{K}}_{\mathrm{b}}=\frac{0.18\times 98\times 60}{2000} \\ {\mathrm{K}}_{\mathrm{b}}=0.5292=\frac{0.002 \times (373{)}^2}{\mathrm{l}} \\ \therefore \mathrm{l}=\frac{0.002\times 373\times 373}{0.5292}=525.8 \mathrm{cal}/\mathrm{g} \end{gathered}$$

(a)

 $$\begin{gathered} ∆\mathrm{T}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}\text{'} \\ \mathrm{m}\text{'}=\frac{∆\mathrm{T}}{{\mathrm{K}}_{\mathrm{f}}}=\frac{0.2}{1.86}=0.1075 \\ 0.1075=\frac{18}{180}\times \frac{1000}{\mathrm{X}} \\ \mathrm{X}=\frac{100}{0.1.75}=930.2 \mathrm{g} \end{gathered}$$

At -0.2$°\mathrm{C}$, the solution has 930.2 g of solvent(water)

$\therefore$amount of ice separated out=1000-930=70 g

(a) Partial pressure of ethane over a saturated solution = 1 bar.

Mass of ethane in the aturated solution at 1 bar = 6.56 X 10^-2 g

Mass of ethane in the solution = 5.00 X 10^-2 g

Partial pressure of ethane gas = ?

According to the Henry's law,

Partial pressure of the gas =K_H X Mole fraction of the gas in solution

So, $1\mathrm{bar}\propto {\mathrm{K}}_{\mathrm{H}}\times 6.56\times {10}^{-2} \mathrm{g}$

and $\mathrm{p}\propto {\mathrm{K}}_{\mathrm{H}}\times 5.00\times {10}^{-2}\mathrm{g}$

So, $\frac{\mathrm{p}}{1 \mathrm{bar}}=\frac{{\mathrm{K}}_{\mathrm{H}}\times 5.00\times {10}^{-2}\mathrm{g}}{{\mathrm{K}}_{\mathrm{H}}\times 6.56\times {10}^{-2} \mathrm{g}}$

or p=$\frac{5.00}{6.56}\times 1 \mathrm{bar}=0.76 \mathrm{bar}$

(b) Let mass of the solution = 100g

Mass of the solute = 2g

Mass of the solvent = 100g-2g= 98g

So, amount of solvent,${\mathrm{n}}_{\mathrm{w}}=\frac{98\mathrm{g}}{{\displaystyle \raisebox{1ex}{$18\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}=5.44 \mathrm{mol}$

Amount of solute, ${\mathrm{n}}_{\mathrm{s}}=\frac{2\mathrm{g}}{\mathrm{M}}$

Then, Mole fraction of the solute, ${\mathrm{X}}_{\mathrm{s}}=\frac{{\displaystyle \raisebox{1ex}{$2\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{M}$}\right.}}{5.44 \mathrm{mol}}$

(assuming solution to be dilute)

From Raoult's law, ${\mathrm{X}}_{\mathrm{s}}=\frac{∆\mathrm{p}}{\mathrm{p}°}=\frac{0.004 \mathrm{bar}}{1 \mathrm{bar}}=0.004$

Therefore,$\frac{2\mathrm{g}}{\mathrm{M}\times 5.44 \mathrm{mol}}=0.004$

or M=$\frac{2\mathrm{g}}{5.44 \mathrm{mol} \times 0.004}=91.9 \mathrm{g} {\mathrm{mol}}^{-1}$

(b) Shows negative deviation from Raoult's law.

If the attraction between different molecules, for example between HCl and H₂O molecules, it is stronger, the escaping tendency from the solution to the vapour phase will be smaller. As a result the partial vapour pressure will be smaller than predicted by Raoult's law and the system exhibits a negative deviation.

(a) $$\begin{gathered} ∆\mathrm{T}={\mathrm{K}}_{\mathrm{f}}^{\text{'}}\times \mathrm{molality}=1.86\times 1=1.86 \\ \therefore \mathrm{Solution} \mathrm{starts} \mathrm{freezing} \mathrm{at} -1.86°\mathrm{C}. \\ \mathrm{Thus}, \mathrm{on} \mathrm{cooling} \mathrm{upto}-3.534°\mathrm{C}, \mathrm{freezing} \mathrm{continues}. \\ \mathrm{Let} \mathrm{molality} \mathrm{of} \mathrm{solution} \mathrm{left} \mathrm{at}-3.534 \mathrm{be} \mathrm{m}\text{'} \\ ∆\mathrm{T}={\mathrm{K}}_{\mathrm{f}}^{\text{'}}\times \mathrm{m}\text{'}; \mathrm{m}\text{'}=\frac{3.534}{1.86}=1.9 \end{gathered}$$

Initialy 1000 gm solvent contains 342 gm sucrose 1342 gm solution contain 342 gm sucrose

1000 gm solutions contain =$\frac{342\times 1000}{1342}\mathrm{sucrose}$

                                      = 254.84 gm sucrose

Finally, amount of water = 1000-254.84=745.16 gm

Since sucrose remains same in solution before and after freezing $\therefore 1.9\times 342 \mathrm{gm}$sucrose is in 1000 gm water 254.84 gm sucrose is in $\frac{1000\times 254.84}{1.9\times 342} \mathrm{gm}$ = 392.18 gm water.

Thus, wt. of ice seperated out = 745.16-392.18=352.98 gm.

(a) No. of partial/ion

BaCl₂ = 3, NaCl = 2, glucose=1

So, order BaCl₂ >NaCl > Glucose

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{b}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{b}}\times \mathrm{m} \\ 1.518=\mathrm{i}\times 2.53\times 1 \\ \mathrm{i}=\frac{1.518}{2.53}=0.6 \\ -\frac{\mathrm{\alpha }}{2}=-0.4 \\ \frac{\mathrm{\alpha }}{2}=0.8 \\ \% \mathrm{association} = 80\% \end{gathered}$$